%I #42 Oct 21 2024 04:29:43
%S 0,0,0,0,0,1,3,7,14,25,42,66,99,143,200,273,364,476,612,775,969,1197,
%T 1463,1771,2125,2530,2990,3510,4095,4750,5481,6293,7192,8184,9275,
%U 10472,11781,13209,14763,16450,18278,20254,22386,24682,27150,29799,32637,35673,38916,42375,46060,49980,54145,58565,63250,68211,73458,79002,84854,91025,97527,104371
%N a(n) = floor(C(n,4)/5).
%C a(n-1) = number of aperiodic necklaces (Lyndon words) with 5 black beads and n-5 white beads.
%D J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 147.
%H Vincenzo Librandi, <a href="/A011795/b011795.txt">Table of n, a(n) for n = 0..1000</a>
%H D. J. Broadhurst, <a href="http://arXiv.org/abs/hep-th/9604128">On the enumeration of irreducible k-fold Euler sums and their roles in knot theory and field theory</a>, arXiv:hep-th/9604128, 1996.
%H <a href="/index/Lu#Lyndon">Index entries for sequences related to Lyndon words</a>
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1,1,-4,6,-4,1).
%F G.f.: x^5*(1+x^3)/((1-x)^3*(1-x^2)*(1-x^5)) = x^5*(1-x+x^2)/((1-x)^5*(1+x+x^2+x^3+x^4)).
%F a(n) = floor(binomial(n+1,5)/(n+1)). - _Gary Detlefs_, Nov 23 2011
%p seq(floor(binomial(n,4)/5), n=0.. 70); # _Zerinvary Lajos_, Jan 12 2009
%t CoefficientList[Series[x^5(1+x^3)/((1-x)^3(1-x^2)(1-x^5)),{x,0,70}],x] (* _Vincenzo Librandi_, Jun 19 2012 *)
%t CoefficientList[Series[x^4/5 (1/(1-x)^5-1/(1- x^5)),{x,0,70}],x] (* _Herbert Kociemba_, Oct 16 2016 *)
%o (Magma) [Floor(Binomial(n+1,5)/(n+1)): n in [0..70]]; // _Vincenzo Librandi_ Jun 19 2012
%o (PARI) a(n)=binomial(n,4)\5 \\ _Charles R Greathouse IV_, Oct 07 2015
%o (SageMath) [binomial(n,4)//5 for n in range(71)] # _G. C. Greubel_, Oct 20 2024
%Y Cf. A000031, A001037, A011915, A051168.
%Y Same as A051170(n+1).
%Y A column of triangle A011847.
%K nonn,easy,changed
%O 0,7
%A _N. J. A. Sloane_, _David Broadhurst_